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Talk:Bashicu matrix system
is there proofs that this is stronger than any nth-order arithmetic? :V Fluoroantimonic Acid (talk) 17:35, January 8, 2016 (UTC) :No. It is stronger than just a "special type of" nth-order logic. �� Fish fish fish ... �� 22:56, January 8, 2016 (UTC) ::And what is this "special type of" nth-order logic we talk about? LittlePeng9 (talk) 08:07, January 9, 2016 (UTC) :::It is a written in the proof of calculation termination (it is written in special terminology and it is difficult to translate into English... I hope some Japanese can translate it into English). For example, calculation of 2-row matrix can be expressed with 2nd order logic. Bashicu matrix does not cover "all of" 2nd order logic (it means that it does not diagonize over the entire 2nd order logic), but special "kind" or "type" of 2nd order logic, as defined in the algorithm. �� Fish fish fish ... �� 08:37, January 9, 2016 (UTC) :::In the proof it is written "n行でn階述語論理のある体系（n階のカインド）を対角化した強さとなる。" Literally it says n-row matrix diagonizes a special type of n-th order logic (n-th order kind). I am not sure if the word "n-th order kind" is a familiar usage of terminology, but anyway in the proof the word "hight of kind" is used. �� Fish fish fish ... �� 08:46, January 9, 2016 (UTC) :::I posted the Japanese BBS for help. Let's wait if some Japanese googlogist who knows logic better than me can explain in more detail... �� Fish fish fish ... �� 09:00, January 9, 2016 (UTC) :::Now the one who posted the proof starts to write more detailed version of the proof. I am asking him or her to make English version of the proof. �� Fish fish fish ... �� 17:05, January 10, 2016 (UTC) ::: ja:ユーザー:黒羽カフカ "バシク行列の計算可能性の証明はもうしばらくお待ちください。" means "Please wait for a while until I rewrite the proof of computability of Bashicu matrix" �� Fish fish fish ... �� 17:02, January 18, 2016 (UTC) ::::Thanks for keeping us updated! Deedlit11 (talk) 16:21, January 20, 2016 (UTC) In Rule 3-2, what are \(L\) and what is \(m\)? While editing I have added "by choice of \(L\)" as I believe that is the case, but I want that clarified. Also, how is \(\Delta\) ever used? It is defined in Rule 3-3, but doesn't appear elsewhere. LittlePeng9 (talk) 20:23, January 8, 2016 (UTC) :From the example, it appears that Rule 3-3 is meant to specify \(\Delta\), not \(N\). I'm assuming that this is correct, and changing it. ~εmli 22:37, January 8, 2016 (UTC) :Looks like \(L\) is \(S_1\). Deedlit11 (talk) 02:16, January 9, 2016 (UTC) :Also, \(m\) is the number of rows in the matrix, or in other words the length of each vector. Deedlit11 (talk) 02:22, January 9, 2016 (UTC) I don't think \(\begin{pmatrix}0&1&2&3&2&3 \\ 0&1&2&3&2&3 \\ 0&1&1&1&1&1\end{pmatrix}\) reach the limit of Taranovsky's notation. Maybe it just reach the limit of the "Degree of reflection" one. {hyp/^,cos} (talk) 14:32, April 12, 2016 (UTC) Column comparison Is the "<" relationship on columns a strict total order? {hyp/^,cos} (talk) 10:59, April 12, 2016 (UTC) : This order isn't total - for example, \((1,2)\) and \((2,1)\) are incomparable. LittlePeng9 (talk) 20:07, April 12, 2016 (UTC) Something wrong happens Something wrong happens at (0,0,0)(1,1,1)(2,0,0)(1,1,0)(2,1,0)(3,1,0)n. When we attemp to solve \(\begin{pmatrix} 0&1&2&1&2&3&2&3&\cdots&2&3&2&3 \\ 0&1&0&1&1&0&1&0&\cdots&1&0&1&1 \\ 0&1&0&0&0&0&0&0&\cdots&0&0&0&0 \end{pmatrix}n\) where there are m \(\begin{matrix}2&3 \\ 1&0 \\ 0&0\end{matrix}\)'s, the sequence for the next step will start from \(\begin{pmatrix} 0&1&2&1&2&3&2&3&\cdots&2&3&2&3 \\ 0&1&0&1&1&0&1&0&\cdots&1&0&1&1 \\ 0&1&0&0&0&0&0&0&\cdots&0&0&0&0 \end{pmatrix}\) where there are m+1 \(\begin{matrix}2&3 \\ 1&0 \\ 0&0\end{matrix}\)'s, and the "(3,0,0)(2,1,0)(3,1,0)" is a part of \(B(1)\). So (0,0,0)(1,1,1)(2,0,0)(1,1,0)(2,1,0)(3,1,0)n reduces to (0,0,0)(1,1,1)(2,0,0)(1,1,0)(2,1,0)(3,0,0)(2,1,0)(3,1,0)...n, where (0,0,0)(1,1,1)(2,0,0)(1,1,0)(2,1,0)(3,0,0)(2,1,0)(3,1,0)n1 reduces to (0,0,0)(1,1,1)(2,0,0)(1,1,0)(2,1,0)(3,0,0)(2,1,0)(3,0,0)(2,1,0)(3,1,0)...n1, where (0,0,0)(1,1,1)(2,0,0)(1,1,0)(2,1,0)(3,0,0)(2,1,0)(3,0,0)(2,1,0)(3,1,0)n2 reduces to (0,0,0)(1,1,1)(2,0,0)(1,1,0)(2,1,0)(3,0,0)(2,1,0)(3,0,0)(2,1,0)(3,0,0)(2,1,0)(3,1,0)...n2, and so on (n < n1 < n2 < ...). So we can never solve (0,0,0)(1,1,1)(2,0,0)(1,1,0)(2,1,0)(3,1,0), and the same as (0,0,0)(1,1,1)(2,0,0)(1,1,0)(2,2,0), (0,0,0)(1,1,1)(2,0,0)(1,1,0)(2,2,1), (0,0,0)(1,1,1)(2,0,0)(1,1,1), (0,0,0)(1,1,1)(2,0,0)(2,0,0), (0,0,0)(1,1,1)(2,0,0)(3,0,0), (0,0,0)(1,1,1)(2,0,0)(3,1,0), (0,0,0)(1,1,1)(2,0,0)(3,1,1), (0,0,0)(1,1,1)(2,1,0), (0,0,0)(1,1,1)(2,1,1), (0,0,0)(1,1,1)(2,2,0), (0,0,0)(1,1,1)(2,2,1) and (0,0,0)(1,1,1)(2,2,2). {hyp/^,cos} (talk) 10:53, April 28, 2016 (UTC) :Yes, it does appear that some of the possible notations do not reduce. However, the system may still be worthwhile googologically; it does seem that some of the notations do reduce, and can get quite large (e.g. (0,0)(1,1)...(n,n)). It would be very useful to find a condition or conditions such that notations that satisfied the condition always reduced. Unfortunately, there is a language barrier between us and the Japanese folks working on the problem. Deedlit11 (talk) 07:19, April 29, 2016 (UTC) The definition was changed. For a maximal j which hold \(\forall i 　 N=(3,1,0) Δ=(0,0,0) (0,0,0)(1,1,1)(2,0,0)(1,1,0)<(2,1,0)>(3,1,0)　　N=(2,1,0) Δ=(1,0,0) (0,0,0)(1,1,1)(2,0,0)<(1,1,0)>(2,1,0)(3,1,0)　　N=(1,1,0) Δ=(2,0,0) (0,0,0)(1,1,1)<(2,0,0)>(1,1,0)(2,1,0)(3,1,0)　　N=(1,1,0) Δ=(2,0,0) (0,0,0)<(1,1,1)>(2,0,0)(1,1,0)(2,1,0)(3,1,0)　　N=(1,1,0) Δ=(2,0,0) <(0,0,0)>(1,1,1)(2,0,0)(1,1,0)(2,1,0)(3,1,0) N=(0,0,0) Δ=(3,0,0) If \(x_j=0\) , a matrix continued from the sequence is a bad part. Thank you for point out mistake, and sorry for my poor English. --KurohaKafka (talk) 22:16, May 12, 2016 (UTC) Another interpret \(x_1\) of sequence N replace \(y_1\) (0,0,0)(1,1,1)(2,0,0)(1,1,0)(2,1,0)<(3,1,0)> 　 N=(3,1,0) Δ=(0,0,0) ... <(0,0,0)>(1,1,1)(2,0,0)(1,1,0)(2,1,0)(3,1,0) N=(0,1,0) Δ=(3,0,0) --KurohaKafka (talk) 03:39, May 13, 2016 (UTC) Oh...what's the new definition of the \(<\) relation on columns? I didn't get it from your words above. {hyp/^,cos} (talk) 15:47, May 13, 2016 (UTC) Another interpretation is mistake. The new definition doesn't use the \(<\) on columns, make sure that \(x_j=0\) or not instead. --KurohaKafka (talk) 12:09, May 14, 2016 (UTC) Okey...but what's the new definition of Rule 3-1 and Rule 3-2? {hyp/^,cos} (talk) 15:09, May 14, 2016 (UTC) Yes, I would like to understand the new rule but we need to see how to compute the next matrix. Deedlit11 (talk) 16:42, May 16, 2016 (UTC) New Rule 3-1 and Rule 3-2 Firstly, \(N=S_n\) and \(\Delta=Z\). \(i\) and \(j\) are as mentioned above. Compare \(N\) with \(S_{n-1}\), and replace \(x_i\) with \(y_i\). If there is no possible value of \(i\) (\(j=0\)), \(N\) does not change. Add \(x_i-y_i\) to \(\Delta_i\). If \(x_j=0\), bad part is \(S_{n-1}\). Otherwise, compare new \(N\) with \(S_{n-2}\). Repeat this until you find \(x_j\) which equal to 0. If not found, replace \(S_n\) with \(Z\). If \(x_j=0\) when you calculate in \(S_k\), bad part is \(S_k\frown S_{k+1}\frown\cdots\frown S_{n-1}\). Rule 3-3 and Rule 3-4 are unchanged. e.g. (0,0,0)(1,1,1)(2,0,0)(1,1,0)(2,1,0)(3,1,0) Bad part is (0,0,0)(1,1,1)(2,0,0)(1,1,0)(2,1,0). \(\Delta\)=(3,0,0) B(1)=(3,0,0)(4,1,1)(5,0,0)(4,1,0)(5,1,0) B(2)=(6,0,0)(7,1,1)(8,0,0)(7,1,0)(8,1,0) …… So (0,0,0)(1,1,1)(2,0,0)(1,1,0)(2,1,0)(3,1,0) is expanded to (0,0,0)(1,1,1)(2,0,0)(1,1,0)(2,1,0)(3,0,0)(4,1,1)(5,0,0)(4,1,0)(5,1,0)... --KurohaKafka (talk) 12:26, May 17, 2016 (UTC) :Ahh...still unclear for me. What's \(i\), \(j\), \(x_i\) and \(y_i\)? {hyp/^,cos} (talk) 23:07, May 17, 2016 (UTC) :\(S_k=(y_0,y_1,\cdots,y_n)\quad N=(x_0,x_1,\cdots,x_n)\) :e.g. :\(S_k=(1,1,1)\quad N=(2,2,1)\) :\(y_0=1\quad y_1=1\quad y_2=1\) and \(x_0=2\quad x_1=2\quad x_2=1\) :\(j=2\ i=0\ \text{or}\ 1\) and \(x_j=1\) :Program is here. バシク行列数(Bashicu matrix number) :--KurohaKafka (talk) 12:36, May 20, 2016 (UTC) :But why the final \(\Delta=(3,0,0)\) in the (0,0,0)(1,1,1)(2,0,0)(1,1,0)(2,1,0)(3,1,0) example? :After 4 steps we've scan to the (1,1,1) in matrix (0,0,0)(1,1,1)(2,0,0)(1,1,0)(2,1,0)(3,1,0), and now \(N=(1,1,0)\) and \(\Delta=(2,0,0)\). :The next step is comparing (0,0,0) with N. We get \(y_0